Range of a Function: Find h(x)

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To find the range of the function h(x) = sqrt(25 + (x - 3)²), it's important to analyze the expression inside the square root. The term (x - 3)² is always non-negative, meaning the minimum value of the radicand is 25 when x = 3. Therefore, the range of h(x) is y ≥ 5, as the square root of 25 is 5. The discussion highlights that expanding the squared term can lead to confusion about the non-negativity of the radicand. Understanding the function's structure clarifies the range effectively.
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Homework Statement



Find the range of h

Homework Equations



h(x) = sqrt(25 + (x - 3)2)

The Attempt at a Solution



I factored out the (x-3)2 and simplified to get

sqrt(x2 - 6x + 34)

I was trying to figure out the domain first, knowing that x2 - 6x >= -34 in order for the number inside the sqrt to be non-negative.. I don't know, maybe I'm missing something really basic here. Help? :)
 
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Codester09 said:

Homework Statement



Find the range of h

Homework Equations



h(x) = sqrt(25 + (x - 3)2)

The Attempt at a Solution



I factored out the (x-3)2 and simplified to get

sqrt(x2 - 6x + 34)
Let's get the terminology straight. You expanded (x-3)2 ; it was already factored.
Codester09 said:
I was trying to figure out the domain first, knowing that x2 - 6x >= -34 in order for the number inside the sqrt to be non-negative.. I don't know, maybe I'm missing something really basic here. Help? :)
The best thing, IMO, was to leave the radicand in its given form, 25 + (x-3)2. Looking at that as its own function, what is the range of this function? That will tell you a lot about the range of h(x).
 
Well the range of 25 + (x - 3)2 is y >= 25, right? So, the range of sqrt(25 + (x - 3)2 is y >= 5?

Yea, I think expanding the (x - 3)2 term messed me up. I was thinking that i was possible for the radicand to be a negative number.. ugh. Stupid mistake.

Thanks a lot for the help.
 
Also, just to let you know...

x^2-6x has a minimum of -9. It will never be less than -34.
 

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